Q9. - Algebra 1

Question

Solve. Assume that $y$ varies inversely as $x$.  

 If $y=8$ when $x=4$ finding $x$ when $y=2$. 

Step-by-Step Solution

Answer

When $y=2,x=16$.

Step 1. State the concept.

Relationship of the form $xy=k$ or $y=\frac{k}{x}$ where $x,y\ne 0$ for some nonzero constant $k$ is known as inverse variation i.e., $y$ varies inversely as $x$.

Product Rule for Inverse Variations – If $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are solutions of an inverse variation, then $x_1{y}_1=x_2{y}_2$.

Step 2. Apply Product rule for Inverse Variations.

To find $y$ when $y=2$, substitute 4 for $x_1$, 8 for $y_1$, and 2 for $y_2$ into the equation $x_1{y}_1=x_2{y}_2$.

 $$\begin{gathered} x_1{y}_1=x_2{y}_2 \\ \left(4\right)\times 8=x_2\left(2\right) \end{gathered}$$

Step 3. Simplify for x 2 .

Further, simplify by dividing both sides by 2.

$$\begin{gathered} \left(4\right)\times 8=x_2\left(2\right) \\ 32=2x_2 \\ x_2=\frac{32}{2} \\ x_2=16 \end{gathered}$$ 

Therefore, $x=16$, when $y=2.$.